Find All $(x,y,z) \in R$ such that :
$\begin{align*} x^2+y^2+xy &= 37 \\ x^2+z^2+zx &= 28 \\ y^2+z^2+yz &= 19 \end{align*}$
My approach is as follows:
I noticed that these expressions are in A.P. And I 've also showed that $x,y,z$ are also in A.P. i.e. $2y=x+z$. But from here, I can't find a way to solve this system of equations.
We have $$x^2+y^2+xy=37\tag1$$ $$x^2+z^2+zx=28\tag2$$ $$y^2+z^2+yz=19\tag3$$ From $(1)-(2)$, $$(y-z)(x+y+z)=9\tag4$$ From $(2)-(3)$, $$(x-y)(x+y+z)=9\tag5$$
From $(4)(5)$, we have $$y-z=x-y\iff 2y=x+z\tag6$$
From $(5)(6)$, $$(x-y)\cdot 3y=9\iff x=y+\frac 3y\tag7$$
Using $(7)$ to eliminate $x$ from $(1)$, $$\begin{align}&\left(y+\frac 3y\right)^2+y^2+\left(y+\frac 3y\right)y=37\\&\Rightarrow 3y^4-28y^2+9=0\\&\Rightarrow (y-3)(y+3)(3y^2-1)=0\\&\Rightarrow y=\pm 3,\quad\pm \frac{1}{\sqrt 3}\\&\end{align}$$
Thus, the answer is $$(x,y,z)=(\pm 4,\pm 3,\pm 2),\left(\pm\frac{10}{3}\sqrt 3,\pm\frac{\sqrt 3}{3},\mp\frac{8}{3}\sqrt 3\right)$$